Question

Differential Equations - Reduction of order

Answer 1

With reduction of order, we assume a solution of the form $y_2=zy_1=ze^{2x}$, with $z=z(x)$. Then

${y_2}'=(z'+2z)e^{2x}$
${y_2}''=(z''+4z'+4z)e^{2x}$

and substituting into the ODE gives

$x(z''+4z'+4z)e^{2x}-(2x+1)(z'+2z)e^{2x}+2ze^{2x}=0$
$x(z''+4z'+4z)-(2x+1)(z'+2z)+2z=0$
$xz''+(2x-1)z'=0$

Let $\xi(x)=z'(x)$, so that $\xi'=z''$. This gives the linear ODE

$x\xi'+(2x-1)\xi=0$

This equation is also separable, so you can write

$\dfrac{\xi'}{\xi}=\dfrac{1-2x}x$

Integrating both sides with respect to $x$ gives

$\ln|\xi|=-2x+\ln x+C_1$
$\xi=C_1xe^{-2x}$

Next, solve $z'=\xi$ for $z$ by integrating both sides again with respect to $x$.

$z'=\xi$
$\implies z=\displaystyle\int C_1xe^{-2x}\,\mathrm dx$
$\implies z=C_1e^{-2x}(2x+1)+C_2$

And finally, solve for $y_2$.

$y_2=zy_1=C_1(2x+1)+C_2e^{2x}$

and note that $y_1$ is already taken into account as part of $y_2$, so this is the general solution to the ODE.